Unitary equivalence between Hermitian operators

Question

Take two (non-zero) Hermitian operators $A$ and $B$. I want to proof that there exists no unitary operator $W$ such that: $$W^{\dagger}AW = A + B$$ For my research I proved this for some specific case that involves symmetries. However I would like to know whether there is a general proof of this statement. If not, are there any conditions on $A$ and $B$ so it is true.
So far I tried solving it with properties of the trace and unitary gates: \begin{align} Tr(A) = Tr(A) + Tr(B) &\rightarrow Tr(B) = 0 \\ Tr(A^2) = Tr(A^2) + Tr(B^2) + 2\cdot Tr(AB) &\rightarrow Tr(B^2) + 2\cdot Tr(AB) = 0 \end{align} And so on. But this does not give me any contradiction e.g. $B = 0$ or something
Hence, my question is can $A$ and $A+B$ be unitarily equivalent?

Edit: included the non-zero condition

Answer 1

The time evolution of the position operator of a free particle in quantum mechanics (Hamiltonian $H=P^2/2m$) is a nice counterexample:

$ e^{iHt/\hbar} X e^{-iHt/\hbar} = X + P t/m$,

with $A=X$ (position operator at time $t=0$), $B= P t/m$ (momentum operator $P$ times $t/\hbar$) and $W= \exp(-iHt/\hbar)$ (time evolution operator).

Answer 2

Here is a simple counter-example: $$e^{iaP/\hbar}Xe^{-iaP/\hbar}=X+a$$ where $X$ is the position operator, $P$ is the momentum operator, and $a$ is a constant.

Answer 3

A nice counterexample has already been presented.

More generally, for any $W$ of the form $e^{X}$, where $X$ is an element of a Lie algebra, it follows from an identity related to the BCH formula that

$e^X A e^{-X} = \sum_{n=0}^{\infty}\frac{[(X)^n, A]}{n!}$.

So in such cases (as in the examples) the statement will clearly fail if $A$ and $X$ do not commute, and their commutator is also Hermitian.